Fibonacci
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12732 | Accepted: 9060 |
Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is
.
Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
0 9 999999999 1000000000 -1
Sample Output
0 34 626 6875
Hint
As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by
.
Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:
.
Source
1 //快速幂矩阵
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 using namespace std;
6 struct Mat{
7 int mat[2][2];
8 };
9 const int Mod = 10000;
10 Mat operator *(Mat a, Mat b)
11 {
12 Mat c;
13 memset(c.mat,0,sizeof(c.mat));
14 for(int i = 0; i < 2; i++)
15 {
16 for(int j = 0; j < 2; j++)
17 {
18 for(int k = 0; k < 2; k++)
19 {
20 c.mat[i][j] = (c.mat[i][j]+(a.mat[i][k]*b.mat[k][j])%Mod)%Mod;
21 }
22 }
23 }
24 return c;
25 }
26 Mat multi(int n)
27 {
28 Mat c;
29 memset(c.mat,0,sizeof(c.mat));
30 c.mat[0][0] = c.mat[1][1] = 1;
31 Mat a;
32 memset(a.mat,0,sizeof(a.mat));
33 a.mat[0][0] = a.mat[0][1] = a.mat[1][0] = 1;
34 while(n)
35 {
36 if(n&1) c = c*a;
37 a = a*a;
38 n>>=1;
39 }
40 return c;
41 }
42 int main()
43 {
44 int n;
45 while(~scanf("%d",&n))
46 {
47 if(n==-1) return 0;
48 Mat ans = multi(n);
49 printf("%d\n",ans.mat[0][1]);
50 }
51 return 0;
52 }